# Use the Laplace transform to solve the given differential equation subject to the indicated initial conditions. y''-7y'+6y =e^t + delta(t-3) + delta(t-5) y(0) = 0 y'(0) = 0

Use the Laplace transform to solve the given differential equation subject to the indicated initial conditions.
${y}^{″}-7{y}^{\prime }+6y={e}^{t}+\delta \left(t-3\right)+\delta \left(t-5\right)$
$y\left(0\right)=0$
${y}^{\prime }\left(0\right)=0$
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Step 1
Here we use table of Laplace transforms and some properties of Laplace transform.
Step 2
${y}^{″}-7{y}^{\prime }+6y={e}^{t}+\delta \left(t-3\right)+\delta \left(t-5\right)$
$y\left(0\right)=0$
${y}^{\prime }\left(0\right)=0$
Apply laplace transform$⇒L\left\{y"\right\}-7L\left\{{y}^{\prime }\right\}+6L\left\{y\right\}=L\left\{{e}^{t}\right\}+L\left\{\delta \left(t-3\right)\right\}+L\left\{\delta \left(t-5\right)\right\}$

$⇒\left({s}^{2}Y\left(s\right)-sY\left(0\right)-{y}^{\prime }\left(0\right)\right)-7\left(sY\left(s\right)-y\left(0\right)\right)+6y\left(s\right)=\frac{1}{\left(s-1\right)}+{e}^{-3s}+{e}^{-5s}$
$⇒\left({s}^{2}-7s+6\right)Y\left(s\right)=\frac{1}{\left(s-1\right)}+{e}^{-3s}+{e}^{-5s}$
$⇒Y\left(s\right)=\frac{1}{\left(s-1\right)\left({s}^{2}-7s+6\right)}+\frac{{e}^{-3s}}{\left({s}^{2}-7s+6\right)}+\frac{{e}^{-5s}}{\left({s}^{2}-7s+6\right)}$
Now apply inverse laplace transform
$⇒{L}^{-1}\left\{Y\left(s\right)\right\}={L}^{-1}\left\{\frac{1}{\left(s-1\right)\left({s}^{2}-7s+6\right)}\right\}+{L}^{-1}\left\{\frac{{e}^{-3s}}{\left({s}^{2}-7s+6\right)}\right\}+{L}^{-1}\left\{\frac{{e}^{-5s}}{\left({s}^{2}-7s+6\right)}\right\}$
Now use paricl and use table of Laplace transforms
$⇒y\left(t\right)=-\frac{1}{6}t\cdot {e}^{t}-\frac{1}{6}{e}^{\left(}t-3\right)\cdot \delta \left(t-3\right)-\frac{1}{6}{e}^{t-5}\cdot \delta \left(t-5\right)$
here we use properties ${L}^{-1}\left\{F\left(s-c\right)\right\}={e}^{ct}\cdot f\left(t\right)$
${L}^{-1}\left\{{e}^{-cs}\cdot F\left(s\right)\right\}=y\left(t\right)\cdot f\left(t\right)$